Your mom is making marmalade by immersed fruit into an aqueous table sucrose solution that freezes at -88°C, what is the osmotic pressure of the fruit at room temperature (25°C)?
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- Your mom is making marmalade by immersed fruit into an aqueous table sucrose solution that freezes at -88°C, what is the osmotic pressure of the fruit at room temperature (25°C)?
NOTE : Solution with -88°C is not possible as the solution is aqueous solution. Hence I believe it is -0.88 °C as I have seen it in other students question.
Hence I'm using the freezing point of solution as -0.88 °C.
Also, Assuming the fruit is made up of sucrose only for simplicity of solution as the molar mass of solute is not given.
Hence assuming the solution to be made up of sucrose only will help in getting molar mass of solute i.e sucrose
Since the solution is aqueous. Hence the depression in freezing point is
depression in freezing point = freezing point of water - freezing point of solution = 0 - (-0.88) = 0.88 oC
Since depression in freezing point is also given by
depression in freezing point = i X Kf X m
where i = Van't Hoff factor = 1 ( since sucrose is a non electrolyte solute)
Kf = depression in freezing point constant = 1.86 oC m-1
m = molality of the solute
Hence substituting the values we get
0.88 = 1 X 1.86 X m
=> m = molality of sucrose = 0.473 mol/Kg
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